Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Prove that : 2(sin 6 θ + cos 6 θ ) – 3 (sin 4 θ + cos 4 θ ) + 1 = 0.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
L.H.S. =
2[(sin 2 θ +cos 2 θ ) 3 – 3 sin 2 θ cos 2 θ (sin 2 θ + cos 2 θ )]
– 3[(sin 2 θ + cos 2 θ ) 2 – 2sin 2 θ cos 2 θ ] + 1
= 2 [1– 3sin 2 θ cos 2 θ ]–3[1–2sin 2 θ cos 2 θ ] + 1
= 0 = R.H.S.
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